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        </a></span></div></div> <div data-v-3ba18f14><main class="page"><div class="page-title" style="display:none;"><h1 class="title">跳表──没听过但很犀利的数据结构</h1> <div data-v-5d8dbdb4><i class="iconfont reco-account" data-v-5d8dbdb4><span data-v-5d8dbdb4>海星</span></i> <!----> <!----> <!----></div></div> <div class="theme-reco-content content__default" style="display:none;"><h1 id="跳表──没听过但很犀利的数据结构"><a href="#跳表──没听过但很犀利的数据结构" class="header-anchor">#</a> 跳表──没听过但很犀利的数据结构</h1> <p>跳表(skip list) 对标的是平衡树(AVL Tree)，是一种 插入/删除/搜索 都是 <code>O(log n)</code> 的数据结构。它最大的优势是原理简单、容易实现、方便扩展、效率更高。因此在一些热门的项目里用来替代平衡树，如 redis, leveldb 等。</p> <h2 id="跳表的基本思想"><a href="#跳表的基本思想" class="header-anchor">#</a> 跳表的基本思想</h2> <p>首先，跳表处理的是有序的链表（一般是双向链表，下图未表示双向），如下：</p> <p><img src="https://lotabout.me/2018/skip-list/linked-list.svg" alt="Linked List"></p> <p>这个链表中，如果要搜索一个数，需要从头到尾比较每个元素是否匹配，直到找到匹配的数为止，即时间复杂度是 $O(n)$。同理，插入一个数并保持链表有序，需要先找到合适的插入位置，再执行插入，总计也是 $O(n)$的时间。</p> <p>那么如何提高搜索的速度呢？很简单，做个索引：</p> <p><img src="https://lotabout.me/2018/skip-list/linked-list-2.svg" alt="Linked List With 2 level"></p> <p>如上图，我们新创建一个链表，它包含的元素为前一个链表的偶数个元素。这样在搜索一个元素时，我们先在上层链表进行搜索，当元素未找到时再到下层链表中搜索。例如搜索数字 <code>19</code> 时的路径如下图：</p> <p><img src="https://lotabout.me/2018/skip-list/linked-list-search-path.svg" alt="Linked List Search Path"></p> <p>先在上层中搜索，到达节点 <code>17</code> 时发现下一个节点为 <code>21</code>，已经大于 <code>19</code>，于是转到下一层搜索，找到的目标数字 <code>19</code>。</p> <p>我们知道上层的节点数目为 $n/2$，因此，有了这层索引，我们搜索的时间复杂度降为了：$O(n/2)$。同理，我们可以不断地增加层数，来减少搜索的时间：</p> <p><img src="https://lotabout.me/2018/skip-list/linked-list-4.svg" alt="Linked List Level 4"></p> <p>在上面的 4 层链表中搜索 <code>25</code>，在最上层搜索时就可以直接跳过 <code>21</code> 之前的所有节点，因此十分高效。</p> <p>更一般地，如果有 kk 层，我们需要的搜索次数会小于 $\frac{n}{2^3} + k$，这样当层数 $k$ 增加到 $\log_2n$时，搜索的时间复杂度就变成了 $logn$。其实这背后的原理和二叉搜索树或二分查找很类似，通过索引来跳过大量的节点，从而提高搜索效率。</p> <h2 id="跳表"><a href="#跳表" class="header-anchor">#</a> 跳表</h2> <p>上节的结构是“静态”的，即我们先拥有了一个链表，再在之上建了多层的索引。但是在实际使用中，我们的链表是通过多次插入/删除形成的，换句话说是“动态”的。上节的结构要求上层相邻节点与对应下层节点间的个数比是 <code>1:2</code>，随意插入/删除一个节点，这个要求就被被破坏了。</p> <p>因此跳表（skip list）表示，我们就不强制要求 <code>1:2</code> 了，一个节点要不要被索引，建几层的索引，都在节点插入时由抛硬币决定。当然，虽然索引的节点、索引的层数是随机的，为了保证搜索的效率，要大致保证每层的节点数目与上节的结构相当。下面是一个随机生成的跳表：</p> <p><img src="https://lotabout.me/2018/skip-list/skip-list.svg" alt="Skip List"></p> <p>可以看到它每层的节点数还和上节的结构差不多，但是上下层的节点的对应关系已经完全被打破了。</p> <p>现在假设节点 <code>17</code> 是最后插入的，在插入之前，我们需要搜索得到插入的位置：</p> <p><img src="https://lotabout.me/2018/skip-list/skip-list-insert-17.svg" alt="Skip List Search Path"></p> <p>接着，抛硬币决定要建立几层的索引，伪代码如下：</p> <div class="language- extra-class"><pre class="language-text"><code>randomLevel()
    lvl := 1
    -- random() that returns a random value in [0...1)
    while random() &lt; p and lvl &lt; MaxLevel do
        lvl := lvl + 1
    return lvl
</code></pre></div><p>上面的伪代码相当于抛硬币，如果是正面（<code>random() &lt; p</code>）则层数加一，直到抛出反面为止。其中的 <code>MaxLevel</code> 是防止如果运气太好，层数就会太高，而太高的层数往往并不会提供额外的性能，一般 $MaxLevel=\log_(\frac{1}{p})n$。现在假设 <code>randomLevel</code> 返回的结果是 <code>2</code>，那么就得到下面的结果。</p> <p><img src="https://lotabout.me/2018/skip-list/skip-list.svg" alt="Skip List"></p> <p>如果要删除节点，则把节点和对应的所有索引节点全部删除即可。当然，要删除节点时需要先搜索得到该节点，搜索过程中可以把路径记录下来，这样删除索引层节点的时候就不需要多次搜索了。</p> <p>显然，在最坏的情况下，所有节点都没有创建索引，时间复杂度为$O(n)$，但在平均情况下，搜索的时间复杂度却是 $O(logn)$，为什么呢？</p> <h2 id="简单的性能分析"><a href="#简单的性能分析" class="header-anchor">#</a> 简单的性能分析</h2> <p>一些严格的证明会涉及到比较复杂的概率统计学知识，所以这里只是简单地说明。</p> <h3 id="每层的节点数目"><a href="#每层的节点数目" class="header-anchor">#</a> 每层的节点数目</h3> <p>上面我们提到 <code>MaxLevel</code>，<a href="ftp://ftp.cs.umd.edu/pub/skipLists/skiplists.pdf">原版论文 </a>中用 <code>L(n)</code> 来表示，要求 <code>L(n)</code> 层有 <code>1/p</code> 个节点，在搜索时可以不理会比 <code>L(n)</code> 更高的层数，直接从 <code>L(n)</code> 层开始搜索，这样效率最高。</p> <p>直观上看，第 l 层的节点中在第 $l+1$ 层也有索引的个数是 $n_{l+1}=n_lp$ 因此第 l 层的节点个数为：</p> <p>$n_l=np^{l−1}$</p> <p>于是代入 $n_{L(n)}=1/p$ 得到 $L(n)=log_{1/p}n$。</p> <h3 id="最高的层数"><a href="#最高的层数" class="header-anchor">#</a> 最高的层数</h3> <p>上面推导到每层的节点数目，直观上看，如果某一层的节点数目小于等于 1，则可以认为它是最高层了，代入 $np^{l−1}=1$ 得到层数 $Lmax=log_{1/p}n+1=L(n)+1=O(logn)$。</p> <p>实际上这个问题并没有直接的解析解，我们能知道的是，当 $n$ 足够大时，最大能达到的层数为 $O(logn)$。</p> <h3 id="搜索的时间复杂度"><a href="#搜索的时间复杂度" class="header-anchor">#</a> 搜索的时间复杂度</h3> <p>为了计算搜索的时间复杂度，我们可以将查找的过程倒过来，从搜索最后的节点开始，一直向左或向上，直到最顶层。如下图，在路径上的每一点，都可能有两种情况：</p> <p><img src="https://lotabout.me/2018/skip-list/skip-list-back-search.svg" alt="Skip List Search Backward"></p> <ol><li>节点有上一层的节点，向上。这种情况出现的概率是 <code>p</code>。</li> <li>节点没有上一层的节点，向左。出现的概率是 <code>1-p</code>。</li></ol> <p>于是，设 <code>C(k)</code> 为反向搜索爬到第 <code>k</code> 层的平均路径长度，则有：</p> <div class="language- extra-class"><pre class="language-text"><code>C(0) = 0
C(k) = p * (情况1) + (1-p) * (情况2)
</code></pre></div><p>将两种情况也用 <code>C</code> 代入，有：</p> <div class="language- extra-class"><pre class="language-text"><code>C(k) = p*(1 + C(k–1)) + (1–p)*(1 + C(k))
C(k) = C(k–1) + 1/p
C(k) = k/p
</code></pre></div><p>上式表明，搜索时，平均在每层上需要搜索的路径长度为 $1/p$，从平均的角度上和我们第一小节构造的“静态”结构相同（p 取 <code>1/2</code>）。</p> <p>又注意到，上小节我们知道跳表的最大层数为 $O(logn)$，因此，搜索的复杂度 $O(logn)/p=O(logn)$。</p> <p>P.S. 这里我们用到的是最大层数，原论文证明时用到的是 $L(n)$，然后再考虑从 $L(n)$ 层到最高层的平均节点个数。这里为了理解方便不再详细证明。</p> <h2 id="小结"><a href="#小结" class="header-anchor">#</a> 小结</h2> <ol><li>各种搜索结构提高效率的方式都是通过空间换时间得到的。</li> <li>跳表最终形成的结构和搜索树很相似。</li> <li>跳表通过随机的方式来决定新插入节点来决定索引的层数。</li> <li>跳表搜索的时间复杂度是 $O(logn)$，插入/删除也是。</li></ol> <p>想到快排(quick sort)与其它排序算法（如归并排序/堆排序）虽然时间复杂度是一样的，但复杂度的常数项较小；跳表的原论文也说跳表能提供一个常数项的速度提升，因此想着常数项小是不是随机算法的一个特点？这也它们大放异彩的重要因素吧。</p> <blockquote><p>来源：https://lotabout.me/2018/skip-list/</p></blockquote> <h2 id="参考"><a href="#参考" class="header-anchor">#</a> 参考</h2> <ul><li><a href="ftp://ftp.cs.umd.edu/pub/skipLists/skiplists.pdf">ftp://ftp.cs.umd.edu/pub/skipLists/skiplists.pdf</a> 原论文</li> <li>https://ticki.github.io/blog/skip-lists-done-right/ skip list 的一些变种、优化</li> <li>https://eugene-eeo.github.io/blog/skip-lists.html skip list 的一些相关复杂度分析</li> <li>http://cglab.ca/~morin/teaching/5408/refs/p90b.pdf skip list cookbook，算是 skip list 各方面的汇总</li> <li><a href="https://juejin.im/entry/59b0eed46fb9a0249471f357" target="_blank" rel="noopener noreferrer">一个可以在有序元素中实现快速查询的数据结构<svg xmlns="http://www.w3.org/2000/svg" aria-hidden="true" x="0px" y="0px" viewBox="0 0 100 100" width="15" height="15" class="icon outbound"><path fill="currentColor" d="M18.8,85.1h56l0,0c2.2,0,4-1.8,4-4v-32h-8v28h-48v-48h28v-8h-32l0,0c-2.2,0-4,1.8-4,4v56C14.8,83.3,16.6,85.1,18.8,85.1z"></path> <polygon fill="currentColor" points="45.7,48.7 51.3,54.3 77.2,28.5 77.2,37.2 85.2,37.2 85.2,14.9 62.8,14.9 62.8,22.9 71.5,22.9"></polygon></svg></a> 包含 skip list 的 C++ 实现</li> <li><a href="http://zhangtielei.com/posts/blog-redis-skiplist.html" target="_blank" rel="noopener noreferrer">Redis内部数据结构详解(6)——skiplist<svg xmlns="http://www.w3.org/2000/svg" aria-hidden="true" x="0px" y="0px" viewBox="0 0 100 100" width="15" height="15" class="icon outbound"><path fill="currentColor" d="M18.8,85.1h56l0,0c2.2,0,4-1.8,4-4v-32h-8v28h-48v-48h28v-8h-32l0,0c-2.2,0-4,1.8-4,4v56C14.8,83.3,16.6,85.1,18.8,85.1z"></path> <polygon fill="currentColor" points="45.7,48.7 51.3,54.3 77.2,28.5 77.2,37.2 85.2,37.2 85.2,14.9 62.8,14.9 62.8,22.9 71.5,22.9"></polygon></svg></a> 图文并茂讲解 skip list，可与本文交叉对照</li> <li>https://www.youtube.com/watch?v=2g9OSRKJuzM MIT 关于 skip list 的课程</li> <li>https://courses.csail.mit.edu/6.046/spring04/handouts/skiplists.pdf MIT 课程讲义</li></ul></div> <footer class="page-edit" style="display:none;"><!----> <!----></footer> <!----> <!----> <!----></main> <!----></div></div></div></div><div class="global-ui"><div class="back-to-ceiling" style="right:1rem;bottom:6rem;width:2.5rem;height:2.5rem;border-radius:.25rem;line-height:2.5rem;display:none;" data-v-db14854a data-v-db14854a><svg t="1574745035067" viewBox="0 0 1024 1024" version="1.1" xmlns="http://www.w3.org/2000/svg" p-id="5404" class="icon" data-v-db14854a><path d="M526.60727968 10.90185116a27.675 27.675 0 0 0-29.21455937 0c-131.36607665 82.28402758-218.69155461 228.01873535-218.69155402 394.07834331a462.20625001 462.20625001 0 0 0 5.36959153 69.94390903c1.00431239 6.55289093-0.34802892 13.13561351-3.76865779 18.80351572-32.63518765 54.11355614-51.75690182 118.55860487-51.7569018 187.94566865a371.06718723 371.06718723 0 0 0 11.50484808 91.98906777c6.53300375 25.50556257 41.68394495 28.14064038 52.69160883 4.22606766 17.37162448-37.73630017 42.14135425-72.50938081 72.80769204-103.21549295 2.18761121 3.04276886 4.15646224 6.24463696 6.40373557 9.22774369a1871.4375 1871.4375 0 0 0 140.04691725 5.34970492 1866.36093723 1866.36093723 0 0 0 140.04691723-5.34970492c2.24727335-2.98310674 4.21612437-6.18497483 6.3937923-9.2178004 30.66633723 30.70611158 55.4360664 65.4791928 72.80769147 103.21549355 11.00766384 23.91457269 46.15860503 21.27949489 52.69160879-4.22606768a371.15156223 371.15156223 0 0 0 11.514792-91.99901164c0-69.36717486-19.13165746-133.82216804-51.75690182-187.92578088-3.42062944-5.66790279-4.76302748-12.26056868-3.76865837-18.80351632a462.20625001 462.20625001 0 0 0 5.36959269-69.943909c-0.00994388-166.08943902-87.32547796-311.81420293-218.6915546-394.09823051zM605.93803103 357.87693858a93.93749974 93.93749974 0 1 1-187.89594924 6.1e-7 93.93749974 93.93749974 0 0 1 187.89594924-6.1e-7z" p-id="5405" data-v-db14854a></path><path d="M429.50777625 765.63860547C429.50777625 803.39355007 466.44236686 1000.39046097 512.00932183 1000.39046097c45.56695499 0 82.4922232-197.00623328 82.5015456-234.7518555 0-37.75494459-36.9345906-68.35043303-82.4922232-68.34111062-45.57627738-0.00932239-82.52019037 30.59548842-82.51086798 68.34111062z" p-id="5406" data-v-db14854a></path></svg></div><!----></div></div>
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